I have three questions related to math?

Question

1. In an algebra class, 21 students are competing to see who can create the most interesting problem involving permutations or combinations. Each student is allowed to submit only one entry. A first prize and a second prize will be awarded. In how many different ways can the instructor award one first prize and one second prize?

2. A field hockey team of nine players elects four of the players as co-captains. How many different combinations of four co-captains are possible.

3. For their family dinner, a local restaurant lets you choose three dishes from column A and two dishes from column B. There are eight items in column A and five items in column B. if no items are repeated, how many different dinners are possible.

Thanks to everyone who answers I really appreciate it.

Answer 1

1. P(21,2) = 21x20 = 420
Here, you are choosing two items (student entries) out of a collection of 21. Order matters, because being chosen for first place is different from being chosen for second place. Another way you could look at it is, for the first prize you have a choice of 21 entries and for the second prize you have a choice of 20 entries. By the fundamental counting principle aka multiplication rule, you connect the two by multiplication.

C(9,4)
Here you are selecting 4 items out of 9, but this time order doesn't matter. When order doen't matter, use combination rather than permutation. This would be 126 ways. Keywords -- committee, co- etc. Beware of the word "combination" in the description of the problem, this can be misleading. Instead look for clues in the description that everyone in the selection will be equal, that order doesn't matter.

C(8,3)xC(5,2) Here, again, order doesn't matter, so we use combinations. WHenver you have an experiment that involves several steps, each of which can be done in various ways, the way the entire experiment can be done is the PRODUCT of the number of ways each step can be done. (this is the fundamental counting principle, loosely stated.) Here, the experiment is chooseing a dinner. First we choose from one column and then we choose from another column. "and then" is a keyword, when you see this in a problem description, tha's a clue that you will likely use the fundamental countinb principle. So we choose from column A, and there are C(8,3) ways = 56 ways to do this. AND THEN we choose from column B, and there are C(5,2) = 10 ways to do this. So there are 56x10 = 560 different dinners possible.

To get these permutations and combinations, you can use the formulas for them (they're in your book or in the links below) or you can use the permutation key and the combination key on your calculator. (I prefer the TI 3x series, especiall the TI-36, for discrete math problems like this, because the graphics calculators have these functions buried under several layers of menus, while the 3x series has dedicated keys._

For more information on this or any other subject put tutorial subject in your search window, then choose from the links a tutorial that suits your needs, one that is written at the right level for you. When I did this for permutations combinations tutorial this came up:
http://math.about.com/gi/dynamic/offsite.htm?zi=1/XJ/Ya&sdn=math&cdn=education&tm=10&gps=66_2_1020_478&f=21&tt=14&bt=0&bts=1&zu=http%3A//www.ping.be/%7Eping1339/tel.htm
http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut57_comb.htm
http://cse.unl.edu/~aallen/Tutorial/tutorial.html
http://library.thinkquest.org/20991/alg2/prob.html

Answer 2

combinations are easier than permutations; combinations is a simple matter of exponents. Permutations is a little harder and involves multiplying n * n-1 (n times)

Question 1 is a permutation because there can only be 1 first prise winner and 1 second prise winner out of 21 possible students and no 2 students can have the same prise. so we multiply
21 * 20 = 420, and since there are only 2 prises there are only 2 numbers to multiply.

Question 2 is a combination and there are 9 players with only 4 players being able to be co-captains so we raise 9 to the 4th power for 6,561 combinations.

Question 3 is a permutation albeit a more complex one. column b permutates while column a does not. The permutations for b is 5 * 4 = 20 now we multiply by 8 to get 160

Answer 3

1. 21 choices for 1st, 20 for second. Since Alice getting 1st place and Bob getting second is different than Bob getting first and Alice getting second, order makes a difference, this is a permutation. The answer is 21P2 or 21 x 20 = 420 ways.

2. This is a combination, because you're selecting groups where everyone in the group is equal. (Alice, Bob, Charlie, Dan) is the same as (Bob, Dan, Alice, Charlie). 9C4 = 9x8x7x6/(4x3x2x1) = 126 ways.

3. This a multiplication of 2 combinations, because you're choosing 3 dishes from 8 (where order makes no difference) and then 2 dishes from 5 (where order again makes no difference).

8C3 x 5C2 = 8x7x6/(3x2x1) x 5x4/(2x1) = 56 x 20 = 560 ways.

I'm assuming you're familiar with the formulas for combinations and permutations, but if not:

nPr = n!/r!

nCr = n!/(r!(n - r)!)

Answer 4

1. 21P2 = 21*20 = 420
And this question won't win a prize.

2. 9C4 = 126

3. (8C3)*(5C2) = 56*10 = 560

Answer 5

no clue.. there's my answer.